# Driving transistors?

Discussion in 'General Electronics Chat' started by danielb33, Nov 4, 2012.

1. ### danielb33 Thread Starter Member

Aug 20, 2012
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I am familiar with driving FETS. Pushing a high voltage pulling a low to make my FET behave in a state according to my needs is easy. I have never had to use transistors. How do people make sure they are getting a specific current or zero current at 100KHz for a switching application? I do not need this now, but am curious about how this works. Controlling states in transistors seems more difficult than FETS. Any other comments are appreciated!

Thanks.

2. ### tshuck Well-Known Member

Oct 18, 2012
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675
You know what FET stands for, right?

Also, are you talking about the mode of operation: linear, cutoff, and saturation?

3. ### Papabravo Expert

Feb 24, 2006
10,021
1,757
About the same really, except you don't have such long delays for charging and discharging the gate capacitance.

1. When the transistor is in cutoff Vbe < 0.7 volts
2. When the transistor is in satuaration Vbe ≈ 0.7V
3. In the linear region Ic ≈ (hfe)(Ib)

Drawing a load line on the collector curves to establish a Q-point is the same exercise as for a FET.

In logic design there are D-FF designers and JK-FF designers. It is the same kind of thing for transistors.

4. ### crutschow Expert

Mar 14, 2008
12,558
3,078
If you use them as switches a bipolar junction transistor (BJT) basically has a current operated input with a base-emitter drop of about 0.7V when on. You just need to apply a current to the base of about 1/10 of the expected collector current to fully turn them on. This current is typically established by a voltage applied to a resistor in series with the base. With no base current they are off.

One disadvantage of BJT's as switches is that they can have a significant storage-time delay when turned off which FET's don't have.

A field-effect transistor (FET) has a voltage operated input, as you know.

Last edited: Nov 8, 2012